\(\int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx\) [69]

Optimal result

Integrand size = 25, antiderivative size = 136 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(3 A-11 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(6 A+13 C) \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {(6 A+13 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )} \]

[Out]

1/7*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+1/35*(3*A-11*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3+1/105*(6*A+13*C)*s
in(d*x+c)/d/(a^2+a^2*cos(d*x+c))^2+1/105*(6*A+13*C)*sin(d*x+c)/d/(a^4+a^4*cos(d*x+c))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3099, 2829, 2729, 2727} \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {(6 A+13 C) \sin (c+d x)}{105 d \left (a^4 \cos (c+d x)+a^4\right )}+\frac {(6 A+13 C) \sin (c+d x)}{105 d \left (a^2 \cos (c+d x)+a^2\right )^2}+\frac {(3 A-11 C) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}+\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[In]

Int[(A + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^4,x]

[Out]

((A + C)*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + ((3*A - 11*C)*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])
^3) + ((6*A + 13*C)*Sin[c + d*x])/(105*d*(a^2 + a^2*Cos[c + d*x])^2) + ((6*A + 13*C)*Sin[c + d*x])/(105*d*(a^4
+ a^4*Cos[c + d*x]))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2829

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 3099

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b
*(A + C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e
+ f*x])^(m + 1)*Simp[a*A*(m + 1) - a*C*m + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C},
x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {\int \frac {-a (3 A-4 C)-7 a C \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2} \\ & = \frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(3 A-11 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(6 A+13 C) \int \frac {1}{(a+a \cos (c+d x))^2} \, dx}{35 a^2} \\ & = \frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(3 A-11 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(6 A+13 C) \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {(6 A+13 C) \int \frac {1}{a+a \cos (c+d x)} \, dx}{105 a^3} \\ & = \frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(3 A-11 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(6 A+13 C) \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {(6 A+13 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.17 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (70 (3 A+4 C) \sin \left (\frac {d x}{2}\right )-175 C \sin \left (c+\frac {d x}{2}\right )+126 A \sin \left (c+\frac {3 d x}{2}\right )+168 C \sin \left (c+\frac {3 d x}{2}\right )-105 C \sin \left (2 c+\frac {3 d x}{2}\right )+42 A \sin \left (2 c+\frac {5 d x}{2}\right )+91 C \sin \left (2 c+\frac {5 d x}{2}\right )+6 A \sin \left (3 c+\frac {7 d x}{2}\right )+13 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{420 a^4 d (1+\cos (c+d x))^4} \]

[In]

Integrate[(A + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^4,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(70*(3*A + 4*C)*Sin[(d*x)/2] - 175*C*Sin[c + (d*x)/2] + 126*A*Sin[c + (3*d*x)/2] +
168*C*Sin[c + (3*d*x)/2] - 105*C*Sin[2*c + (3*d*x)/2] + 42*A*Sin[2*c + (5*d*x)/2] + 91*C*Sin[2*c + (5*d*x)/2]
+ 6*A*Sin[3*c + (7*d*x)/2] + 13*C*Sin[3*c + (7*d*x)/2]))/(420*a^4*d*(1 + Cos[c + d*x])^4)

Maple [A] (verified)

Time = 1.78 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.57

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 \left (3 A -C \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+7 \left (A -\frac {C}{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 A +7 C \right )}{56 a^{4} d}\)	\(78\)
derivativedivides	\(\frac {\frac {\left (A +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (3 A -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (3 A -C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(88\)
default	\(\frac {\frac {\left (A +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (3 A -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (3 A -C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(88\)
risch	\(\frac {2 i \left (105 C \,{\mathrm e}^{5 i \left (d x +c \right )}+175 C \,{\mathrm e}^{4 i \left (d x +c \right )}+210 A \,{\mathrm e}^{3 i \left (d x +c \right )}+280 C \,{\mathrm e}^{3 i \left (d x +c \right )}+126 A \,{\mathrm e}^{2 i \left (d x +c \right )}+168 C \,{\mathrm e}^{2 i \left (d x +c \right )}+42 A \,{\mathrm e}^{i \left (d x +c \right )}+91 C \,{\mathrm e}^{i \left (d x +c \right )}+6 A +13 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(126\)
norman	\(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\left (A +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}+\frac {\left (9 A +5 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}+\frac {\left (27 A +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (31 A +3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{280 a d}+\frac {\left (123 A -31 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{420 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{3}}\)	\(165\)

[In]

int((A+C*cos(d*x+c)^2)/(a+cos(d*x+c)*a)^4,x,method=_RETURNVERBOSE)

[Out]

1/56*tan(1/2*d*x+1/2*c)*((A+C)*tan(1/2*d*x+1/2*c)^6+7/5*(3*A-C)*tan(1/2*d*x+1/2*c)^4+7*(A-1/3*C)*tan(1/2*d*x+1
/2*c)^2+7*A+7*C)/a^4/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.90 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {{\left ({\left (6 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (6 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (39 \, A + 32 \, C\right )} \cos \left (d x + c\right ) + 36 \, A + 8 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

[In]

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*((6*A + 13*C)*cos(d*x + c)^3 + 4*(6*A + 13*C)*cos(d*x + c)^2 + (39*A + 32*C)*cos(d*x + c) + 36*A + 8*C)*
sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) +
a^4*d)

Sympy [A] (verification not implemented)

Time = 1.60 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.31 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\begin {cases} \frac {A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} + \frac {3 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} + \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\left (c \right )}\right )}{\left (a \cos {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate((A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**4,x)

[Out]

Piecewise((A*tan(c/2 + d*x/2)**7/(56*a**4*d) + 3*A*tan(c/2 + d*x/2)**5/(40*a**4*d) + A*tan(c/2 + d*x/2)**3/(8*
a**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d) + C*tan(c/2 + d*x/2)**7/(56*a**4*d) - C*tan(c/2 + d*x/2)**5/(40*a**4*d
) - C*tan(c/2 + d*x/2)**3/(24*a**4*d) + C*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*(A + C*cos(c)**2)/(a*cos(
c) + a)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.29 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

[In]

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(C*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 + 3*A*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7)/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.86 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

[In]

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 + 15*C*tan(1/2*d*x + 1/2*c)^7 + 63*A*tan(1/2*d*x + 1/2*c)^5 - 21*C*tan(1/2*
d*x + 1/2*c)^5 + 105*A*tan(1/2*d*x + 1/2*c)^3 - 35*C*tan(1/2*d*x + 1/2*c)^3 + 105*A*tan(1/2*d*x + 1/2*c) + 105
*C*tan(1/2*d*x + 1/2*c))/(a^4*d)

Mupad [B] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{8\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A-C\right )}{24\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,A-C\right )}{40\,a^4}}{d} \]

[In]

int((A + C*cos(c + d*x)^2)/(a + a*cos(c + d*x))^4,x)

[Out]

((tan(c/2 + (d*x)/2)^7*(A + C))/(56*a^4) + (tan(c/2 + (d*x)/2)*(A + C))/(8*a^4) + (tan(c/2 + (d*x)/2)^3*(3*A -
C))/(24*a^4) + (tan(c/2 + (d*x)/2)^5*(3*A - C))/(40*a^4))/d